Division of Algebraic Expressions
Division of algebraic expressions is the reverse of multiplication. Just as 12 ÷ 3 = 4 because 3 × 4 = 12, dividing algebraic expressions means finding what expression, when multiplied by the divisor, gives the dividend.
In Class 8 NCERT Maths, you learn to divide a monomial by a monomial, a polynomial by a monomial, and a polynomial by a polynomial using factorisation.
The key technique is to factorise both the dividend and the divisor, then cancel common factors. This is similar to simplifying fractions in arithmetic.
Division of algebraic expressions is essential for simplifying algebraic fractions, solving equations, and working with rational expressions in higher classes.
What is Division of Algebraic Expressions?
Definition: Dividing one algebraic expression by another means finding the quotient such that:
Dividend = Divisor × Quotient + Remainder
In Class 8, most problems result in exact division (remainder = 0).
Three cases of division:
- Monomial ÷ Monomial: Divide coefficients and subtract exponents.
- Polynomial ÷ Monomial: Divide each term of the polynomial by the monomial.
- Polynomial ÷ Polynomial: Factorise both, then cancel common factors.
Division of Algebraic Expressions Formula
Monomial ÷ Monomial:
(a · xᵐ) ÷ (b · xⁿ) = (a/b) · xᵐ⁻ⁿ (where m ≥ n)
Polynomial ÷ Monomial:
(ax² + bx + c) ÷ x = ax + b + c/x
Law of exponents used:
- xᵐ ÷ xⁿ = xᵐ⁻ⁿ (when m > n)
- xᵐ ÷ xᵐ = x⁰ = 1
- xᵐ ÷ xⁿ = 1/xⁿ⁻ᵐ (when m < n)
Derivation and Proof
Why does dividing polynomials by factorisation work?
Step 1: Consider: (6x² + 9x) ÷ 3x.
Step 2: Factorise the dividend: 6x² + 9x = 3x(2x + 3).
Step 3: Now divide: 3x(2x + 3) ÷ 3x = (2x + 3).
This works because division is the inverse of multiplication. If A = B × C, then A ÷ B = C.
Why subtract exponents?
- x⁵ ÷ x² = (x × x × x × x × x) ÷ (x × x) = x × x × x = x³ = x⁵⁻²
This is the law of exponents: xᵐ ÷ xⁿ = xᵐ⁻ⁿ.
Types and Properties
Types of division problems:
1. Monomial ÷ Monomial:
- Example: 12x⁵ ÷ 4x² = 3x³
2. Polynomial ÷ Monomial:
- Divide each term separately.
- Example: (8x³ + 4x² − 2x) ÷ 2x = 4x² + 2x − 1
3. Polynomial ÷ Polynomial (using factorisation):
- Factorise both, cancel common factors.
- Example: (x² − 9) ÷ (x + 3) = (x+3)(x−3) ÷ (x+3) = x − 3
4. Division to simplify algebraic fractions:
- Example: (x² + 5x + 6) / (x + 2) = (x+2)(x+3) / (x+2) = x + 3
Solved Examples
Example 1: Example 1: Monomial ÷ Monomial
Problem: Divide 15x⁴ by 3x².
Solution:
- Coefficients: 15 ÷ 3 = 5
- Variables: x⁴ ÷ x² = x⁴⁻² = x²
- Result = 5x²
Answer: 15x⁴ ÷ 3x² = 5x².
Example 2: Example 2: Monomial ÷ Monomial with two variables
Problem: Divide 20a³b² by 5a²b.
Solution:
- Coefficients: 20 ÷ 5 = 4
- Variables: a³ ÷ a² = a; b² ÷ b = b
- Result = 4ab
Answer: 20a³b² ÷ 5a²b = 4ab.
Example 3: Example 3: Polynomial ÷ Monomial
Problem: Divide (12x³ + 8x² − 4x) by 4x.
Solution:
- 12x³ ÷ 4x = 3x²
- 8x² ÷ 4x = 2x
- −4x ÷ 4x = −1
- Result = 3x² + 2x − 1
Answer: (12x³ + 8x² − 4x) ÷ 4x = 3x² + 2x − 1.
Example 4: Example 4: Polynomial ÷ Monomial with negative coefficient
Problem: Divide (18y⁴ − 27y³ + 9y²) by −3y².
Solution:
- 18y⁴ ÷ (−3y²) = −6y²
- −27y³ ÷ (−3y²) = 9y
- 9y² ÷ (−3y²) = −3
- Result = −6y² + 9y − 3
Answer: (18y⁴ − 27y³ + 9y²) ÷ (−3y²) = −6y² + 9y − 3.
Example 5: Example 5: Polynomial ÷ Polynomial using factorisation
Problem: Divide (x² − 16) by (x + 4).
Solution:
- Factorise: x² − 16 = (x + 4)(x − 4) [difference of squares]
- (x + 4)(x − 4) ÷ (x + 4) = x − 4
Answer: (x² − 16) ÷ (x + 4) = x − 4.
Example 6: Example 6: Division using identity factorisation
Problem: Divide (4x² + 12x + 9) by (2x + 3).
Solution:
- Recognise: 4x² + 12x + 9 = (2x)² + 2(2x)(3) + 3² = (2x + 3)²
- (2x + 3)² ÷ (2x + 3) = 2x + 3
Answer: (4x² + 12x + 9) ÷ (2x + 3) = 2x + 3.
Example 7: Example 7: Factorisation by common factor then division
Problem: Divide (6x³ + 12x² + 6x) by (x² + 2x + 1).
Solution:
- Numerator: 6x³ + 12x² + 6x = 6x(x² + 2x + 1) = 6x(x + 1)²
- Denominator: x² + 2x + 1 = (x + 1)²
- 6x(x + 1)² ÷ (x + 1)² = 6x
Answer: (6x³ + 12x² + 6x) ÷ (x² + 2x + 1) = 6x.
Example 8: Example 8: Factorisation by grouping then division
Problem: Divide (x³ + x² + x + 1) by (x + 1).
Solution:
- Factor by grouping: x³ + x² + x + 1 = x²(x + 1) + 1(x + 1) = (x + 1)(x² + 1)
- (x + 1)(x² + 1) ÷ (x + 1) = x² + 1
Answer: (x³ + x² + x + 1) ÷ (x + 1) = x² + 1.
Example 9: Example 9: Simplifying an algebraic fraction
Problem: Simplify: (x² − 5x + 6) / (x − 2).
Solution:
- Factorise numerator: x² − 5x + 6 = (x − 2)(x − 3)
- (x − 2)(x − 3) / (x − 2) = x − 3
Answer: The simplified expression is x − 3.
Example 10: Example 10: Finding the missing factor
Problem: If the area of a rectangle is (6x² + 11x + 3) and the length is (2x + 3), find the breadth.
Solution:
- Breadth = Area ÷ Length = (6x² + 11x + 3) ÷ (2x + 3)
- Factorise: 6x² + 11x + 3 = (2x + 3)(3x + 1)
- Breadth = (2x + 3)(3x + 1) ÷ (2x + 3) = 3x + 1
Answer: The breadth is (3x + 1).
Real-World Applications
Simplifying Fractions: Algebraic fractions (rational expressions) are simplified by dividing numerator and denominator by their common factors, exactly like numerical fractions.
Solving Equations: Dividing both sides of an equation by an expression is a common step in solving linear and quadratic equations.
Finding Dimensions: If the area/volume and one dimension are given as algebraic expressions, dividing gives the other dimension.
Factorisation Verification: Division can verify factorisation: if P(x) = Q(x) × R(x), then P(x) ÷ Q(x) should give R(x) exactly.
Higher Mathematics: Polynomial long division (Class 9–10) and synthetic division build on this foundation. These techniques are essential in algebra, calculus, and engineering.
Key Points to Remember
- Division of algebraic expressions is the reverse of multiplication.
- Monomial ÷ Monomial: divide coefficients and subtract exponents (xᵐ ÷ xⁿ = xᵐ⁻ⁿ).
- Polynomial ÷ Monomial: divide each term of the polynomial by the monomial.
- Polynomial ÷ Polynomial: factorise both, then cancel common factors.
- Always factorise completely before dividing.
- Sign rules apply: (−) ÷ (−) = (+), (+) ÷ (−) = (−).
- Division by zero is not defined.
- The quotient’s degree = dividend’s degree − divisor’s degree (for exact division).
- Use algebraic identities to factorise when possible.
- Division can be checked: Divisor × Quotient = Dividend (for exact division).
Practice Problems
- Divide 24x⁶ by 8x³.
- Divide (15a⁴b³) by (5a²b²).
- Divide (9x³ − 12x² + 6x) by 3x.
- Divide (x² − 25) by (x − 5).
- Simplify: (x² + 7x + 12) / (x + 3).
- Divide (8x³ − 27) by (2x − 3). [Hint: Use a³ − b³ = (a−b)(a²+ab+b²)]
- If the volume of a cuboid is (6x³ + x² − 2x) and the base area is (2x² + x), find the height.
- Simplify: (4x² − 1) / (2x + 1).
Frequently Asked Questions
Q1. How do you divide a monomial by a monomial?
Divide the coefficients and subtract the exponents. For example, 12x⁵ ÷ 4x² = 3x³.
Q2. How do you divide a polynomial by a monomial?
Divide each term of the polynomial by the monomial separately. For example, (6x² + 9x) ÷ 3x = 2x + 3.
Q3. How do you divide a polynomial by another polynomial?
Factorise both the dividend and the divisor. Cancel the common factors to get the quotient.
Q4. Why do we need to factorise before dividing polynomials?
Factorisation reveals common factors that cancel, making the division straightforward. Without factorisation, polynomial long division would be needed.
Q5. What is the law of exponents for division?
xᵐ ÷ xⁿ = xᵐ⁻ⁿ. For example, x⁵ ÷ x² = x³.
Q6. Can you divide by zero?
No. Division by zero is undefined. If the divisor is an expression like (x − 3), the result is valid only when x ≠ 3.
Q7. How do you check if division is correct?
Multiply the quotient by the divisor. If the result equals the dividend, the division is correct.
Q8. What happens when the exponent of the divisor is greater?
You get a negative exponent: x² ÷ x⁵ = x⁻³ = 1/x³. In Class 8, such divisions are usually avoided.
Q9. Is (x² − 4) ÷ (x − 2) the same as (x + 2)?
Yes, because x² − 4 = (x+2)(x−2), so (x+2)(x−2) ÷ (x−2) = x+2 (provided x ≠ 2).
Q10. What identities help in factorising for division?
Common identities: a² − b² = (a+b)(a−b), a² + 2ab + b² = (a+b)², a² − 2ab + b² = (a−b)², and a³ − b³ = (a−b)(a²+ab+b²).
